MSCA 605 HW 7
Matt
5/10/2020
Question 1
One-Factor Anova Test
H0: All 4 means for the Drugs will be equal HA: At least one of the means for the Drugs will not be equal.
## Call:
## aov(formula = value ~ variable, data = mydrug2)
##
## Terms:
## variable Residuals
## Sum of Squares 698.2 793.6
## Deg. of Freedom 3 16
##
## Residual standard error: 7.042727
## Estimated effects may be unbalanced## Analysis of Variance Table
##
## Response: value
## Df Sum Sq Mean Sq F value Pr(>F)
## variable 3 698.2 232.73 4.6922 0.01553 *
## Residuals 16 793.6 49.60
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1##
## Pairwise comparisons using t tests with pooled SD
##
## data: mydrug2$value and mydrug$variable
##
## Drug1 Drug2 Drug3
## Drug2 1.000 - -
## Drug3 0.165 0.235 -
## Drug4 1.000 1.000 0.012
##
## P value adjustment method: bonferroni##
## Pairwise comparisons using t tests with non-pooled SD
##
## data: mydrug2$value and mydrug$variable
##
## Drug1 Drug2 Drug3
## Drug2 1.00 - -
## Drug3 0.29 0.14 -
## Drug4 1.00 1.00 0.04
##
## P value adjustment method: bonferroni## Tukey multiple comparisons of means
## 95% family-wise confidence level
##
## Fit: aov(formula = value ~ variable, data = mydrug2)
##
## $variable
## diff lwr upr p adj
## Drug2-Drug1 -0.8 -13.543587 11.943587 0.9978542
## Drug3-Drug1 -10.8 -23.543587 1.943587 0.1122136
## Drug4-Drug1 5.6 -7.143587 18.343587 0.6014607
## Drug3-Drug2 -10.0 -22.743587 2.743587 0.1533294
## Drug4-Drug2 6.4 -6.343587 19.143587 0.4959400
## Drug4-Drug3 16.4 3.656413 29.143587 0.0097855The anova test shows that not all of the means are equal. The test produced a p-value less than .05 meaning we would reject the Null and accept the alternate. Drug 3 and Drug 4 appear to have the most significant difference among the different drugs.
Descriptive Statistics
When looking at the descriptive statistics, you can clearly see that the means for Drugs 3 and 4 are very different than those of 1 and 2. 1 and 2 are actually very close together.
##
## Descriptive statistics by group
## group: Drug1
## vars n mean sd median trimmed mad min max range skew kurtosis
## variable* 1 5 1.0 0.00 1 1.0 0.00 1 1 0 NaN NaN
## value 2 5 26.4 8.76 26 26.4 5.93 14 38 24 -0.09 -1.58
## se
## variable* 0.00
## value 3.92
## ------------------------------------------------------------
## group: Drug2
## vars n mean sd median trimmed mad min max range skew kurtosis se
## variable* 1 5 2.0 0.00 2 2.0 0.0 2 2 0 NaN NaN 0.00
## value 2 5 25.6 6.54 28 25.6 8.9 18 34 16 0 -1.98 2.93
## ------------------------------------------------------------
## group: Drug3
## vars n mean sd median trimmed mad min max range skew kurtosis
## variable* 1 5 3.0 0.00 3 3.0 0.00 3 3 0 NaN NaN
## value 2 5 15.6 3.85 16 15.6 2.97 10 20 10 -0.28 -1.72
## se
## variable* 0.00
## value 1.72
## ------------------------------------------------------------
## group: Drug4
## vars n mean sd median trimmed mad min max range skew kurtosis se
## variable* 1 5 4 0 4 4 0.00 4 4 0 NaN NaN 0.00
## value 2 5 32 8 30 32 5.93 22 44 22 0.28 -1.5 3.58Visualization
The boxplot shows that there are 3 different data points that are outliers. It also appears that Drug3 and Drug4 are pretty different from Drug1 and Drug2.
Check Assumptions
Assumption 1: Normality
The Shapiro-Wilk tests show that each of the drugs have normal distributions. Outliers were not removed, so some of the normality may be affected by this, but it appears not too significantly.
##
## Shapiro-Wilk normality test
##
## data: Z
## W = 0.98549, p-value = 0.9617##
## Shapiro-Wilk normality test
##
## data: Z
## W = 0.92221, p-value = 0.5443##
## Shapiro-Wilk normality test
##
## data: Z
## W = 0.97872, p-value = 0.9276##
## Shapiro-Wilk normality test
##
## data: Z
## W = 0.9491, p-value = 0.7308Assumption 2: Check for Outliers
Three outliers were identified. One for Drug1 and two for Drug4.
Multivariate Repeated Measure Analysis
The multivariate repeated measure analysis shows that Drug3 and and Drug4 are significantly different at the p = .05 confidence level. The analysis shows that Drug3’s mean is 10.8 points less than that of Drug1. Drug 4 has a mean that is 5.6 points higher than that of Drug1. Drug2 is very similar to Drug1.
## Linear mixed-effects model fit by maximum likelihood
## Data: mydrug
## AIC BIC logLik
## 123.5902 129.5646 -55.79509
##
## Random effects:
## Formula: ~1 | Subjects
## (Intercept) Residual
## StdDev: 5.670979 2.742262
##
## Fixed effects: value ~ variable
## Value Std.Error DF t-value p-value
## (Intercept) 26.4 3.149603 12 8.382008 0.0000
## variableDrug2 -0.8 1.939072 12 -0.412568 0.6872
## variableDrug3 -10.8 1.939072 12 -5.569675 0.0001
## variableDrug4 5.6 1.939072 12 2.887979 0.0136
## Correlation:
## (Intr) vrblD2 vrblD3
## variableDrug2 -0.308
## variableDrug3 -0.308 0.500
## variableDrug4 -0.308 0.500 0.500
##
## Standardized Within-Group Residuals:
## Min Q1 Med Q3 Max
## -1.53063969 -0.57831784 -0.04002289 0.69521552 1.52978267
##
## Number of Observations: 20
## Number of Groups: 5## Warning: Converting "Subjects" to factor for ANOVA.## $ANOVA
## Effect DFn DFd SSn SSd F p p<.05 ges
## 1 (Intercept) 1 4 12400.2 680.8 72.85664 1.033910e-03 * 0.9398505
## 2 variable 3 12 698.2 112.8 24.75887 1.992501e-05 * 0.4680252
##
## $`Mauchly's Test for Sphericity`
## Effect W p p<.05
## 2 variable 0.1864953 0.4957455
##
## $`Sphericity Corrections`
## Effect GGe p[GG] p[GG]<.05 HFe p[HF] p[HF]<.05
## 2 variable 0.604874 0.0006490326 * 1.078854 1.992501e-05 *Question 2
Boxplots
The box plots show that there are a few outliers. In addition, Versicolor and Virginica (the second two plots) have similar measurements in each category.
Covariances and Correlations
The first portion of the data represents Setosa. The second, versicolor. The third, Virginica. There appears to be much more correlation with Setosa than the other two types of Iris flowers.
## Sepal.Length Sepal.Width Petal.Length Petal.Width
## Sepal.Length 0.12424898 0.099216327 0.016355102 0.010330612
## Sepal.Width 0.09921633 0.143689796 0.011697959 0.009297959
## Petal.Length 0.01635510 0.011697959 0.030159184 0.006069388
## Petal.Width 0.01033061 0.009297959 0.006069388 0.011106122## Sepal.Length Sepal.Width Petal.Length Petal.Width
## Sepal.Length 1.0000000 0.7425467 0.2671758 0.2780984
## Sepal.Width 0.7425467 1.0000000 0.1777000 0.2327520
## Petal.Length 0.2671758 0.1777000 1.0000000 0.3316300
## Petal.Width 0.2780984 0.2327520 0.3316300 1.0000000## [1] 0.3533595## Sepal.Length Sepal.Width Petal.Length Petal.Width
## Sepal.Length 0.26643265 0.08518367 0.18289796 0.05577959
## Sepal.Width 0.08518367 0.09846939 0.08265306 0.04120408
## Petal.Length 0.18289796 0.08265306 0.22081633 0.07310204
## Petal.Width 0.05577959 0.04120408 0.07310204 0.03910612## Sepal.Length Sepal.Width Petal.Length Petal.Width
## Sepal.Length 1.0000000 0.5259107 0.7540490 0.5464611
## Sepal.Width 0.5259107 1.0000000 0.5605221 0.6639987
## Petal.Length 0.7540490 0.5605221 1.0000000 0.7866681
## Petal.Width 0.5464611 0.6639987 0.7866681 1.0000000## [1] 0.08359418## Sepal.Length Sepal.Width Petal.Length Petal.Width
## Sepal.Length 0.40434286 0.09376327 0.30328980 0.04909388
## Sepal.Width 0.09376327 0.10400408 0.07137959 0.04762857
## Petal.Length 0.30328980 0.07137959 0.30458776 0.04882449
## Petal.Width 0.04909388 0.04762857 0.04882449 0.07543265## Sepal.Length Sepal.Width Petal.Length Petal.Width
## Sepal.Length 1.0000000 0.4572278 0.8642247 0.2811077
## Sepal.Width 0.4572278 1.0000000 0.4010446 0.5377280
## Petal.Length 0.8642247 0.4010446 1.0000000 0.3221082
## Petal.Width 0.2811077 0.5377280 0.3221082 1.0000000## [1] 0.1373902Shapiro Tests
The Shapiro tests show that Versicolor and Virginica are not normally distributed using the p = .05 interval. Setosa, although not statistically significant, is nearly not normally distributed. At the p = .05 level, it would be considered normally distributed.
##
## Shapiro-Wilk normality test
##
## data: Z
## W = 0.95878, p-value = 0.07906##
## Shapiro-Wilk normality test
##
## data: Z
## W = 0.93043, p-value = 0.005739##
## Shapiro-Wilk normality test
##
## data: Z
## W = 0.93414, p-value = 0.007955Profile
The graph below shows that the the measurements of the different groups are non-linear. This is supported by the p-values found below.
##
## Data Summary:
## setosa versicolor virginica
## Sepal.Length 5.006 5.936 6.588
## Sepal.Width 3.428 2.770 2.974
## Petal.Length 1.462 4.260 5.552
## Petal.Width 0.246 1.326 2.026## Call:
## pbg(data = iris[, 1:4], group = iris[, 5], original.names = TRUE,
## profile.plot = TRUE)
##
## Hypothesis Tests:
## $`Ho: Profiles are parallel`
## Multivariate.Test Statistic Approx.F num.df den.df p.value
## 1 Wilks 0.04115317 189.92337 6 290 2.395832e-97
## 2 Pillai 0.96909246 45.74852 6 292 2.472886e-39
## 3 Hotelling-Lawley 23.05050400 553.21210 6 288 7.441946e-155
## 4 Roy 23.03969817 1121.26531 3 146 1.477137e-100
##
## $`Ho: Profiles have equal levels`
## Df Sum Sq Mean Sq F value Pr(>F)
## group 2 77.40 38.70 422.4 <2e-16 ***
## Residuals 147 13.47 0.09
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## $`Ho: Profiles are flat`
## F df1 df2 p-value
## 1 4847.465 3 145 3.788058e-145